SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.

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Jasmin Copeland
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1 SETION Evaluation of Dfinit Intgrals Exampl 6.6 W hav usd dfinit intgrals to valuat contour intgrals. It may com as a surpris to larn that contour intgrals and rsidus can b usd to valuat crtain classs of dfinit intgrals that might othrwis prov intractabl. Dfinit Intgrals Involving Trigonomtric Functions ontour intgrals and rsidus can b usful in th valuation of dfinit intgrals of th form b P (cos θ, sin θ) dθ (6.4) a Q(cos θ, sin θ) whr P (cos θ, sin θ) and Q(cos θ, sin θ) ar polynomials in cos θ and sin θ, providd Q is nvr qual to zro. π cos θ dθ. Solution W transform th dfinit intgral to th complx plan by stting z θi and dz i θi dθ iz dθ. Asθ tracs out th valus through π, z tracs out th circl : z onc countrclockwis (Figur 6.5). W rplac cos θ with cos θ ( θi + θi )/ (z + z )/, π cos θ dθ ( ) dz iz i z 4z + dz. z+z Th (ral) dfinit intgral has bn rplacd by a (complx) contour intgral. Bcaus z 4z + whn z (4± 6 4)/ ±, th intgrand z 4z + (z )(z + ) has simpl pols at z ±, only on of which is intrior to. Sinc Rs z 4z +, ] z + lim z (z )(z + ), auchy s rsidu thorm givs π ( dθ i πi )] cos θ π. i, - R z D i, -, + R z Figur 6.5 Figur 6.6 This mthod can b applid to valuat intgrals ovr intrvals othr than, π], but rsidu thory may not b applicabl.

2 58 SETION 6. Exampl 6.7 π/ cos θ dθ. Solution Th transformation z θi transforms th ral intgral into th contour intgral of Exampl 6.6 ovr th quartr circl in Figur 6.6. This tim w us th partial fraction dcomposition of th intgrand to writ π/ dθ i cos θ z dz i 4z + ( z + ) z + dz. In th domain D of Figur 6.6, th intgrand has an antidrivativ, and hnc π/ cos θ dθ i {log φ (z ) log φ (z + ) whr w choos branchs of th logarithm functions with branch cuts φ π/. Thn π/ cos θ dθ i log π/ (i ) log π/ (i + ) log π/ ( ) + log π/ ( + ] ) { i ln ( ( ( ) ++i π Tan ln ( + ( ) ++i ln ( + ) + πi]+ln( ) } i, ))] + ( ))] π Tan } { i ln 8+4 ln 8 4 ln ( + )+ln( ] ) ( ) ( )] } + i π +Tan Tan +. W can bring th invrs tangnts togthr using th following idntity. AB >, ( ) A B Tan A Tan B Tan. +AB Th rsult is π/ cos θ dθ i ln ( ] ) ( +) Whn π +Tan + ( )( ) + +

3 SETION i ln ] 4+ π +Tan ] ) π. 9 Ral Impropr Intgrals Exampl 6.8 Rsidus can also b ffctiv in valuation of impropr intgrals which hav infinit uppr or lowr limits. +x dx. 4 Solution It is fairly clar that wr w to valuat th contour intgral +z 4 dz whr is shown in Figur 6.7a, and wr w to lt R, thn that part of th contour intgral along th ral axis would giv ris to th rquird impropr intgral. Lt us considr this contour intgral thn. G pi /4 p i /4 pi /4 p i /4 -R R > R z -R R > R z Figur 6.7a Figur 6.7b Th intgrand ( + z 4 ) has simpl pols at th four fourth roots of, πi/4, πi/4, 5πi/4, 7πi/4, only th first two of which ar intrior to. L Hôpital s rul (Thorm 5.4) givs ] z πi/4 Rs +z 4,πi/4 lim lim z πi/4 +z 4 z πi/4 4z 4 ( + i). πi/4 8 ] Similarly, Rs +z 4,πi/4 ( i). By auchy s rsidu thorm thn, 8 ] dz πi ( + i)+ ( i) π. +z4 8 8 Suppos w now divid into a smicircular part Γ and a straight lin part (Figur 6.7b). Thn π R +x dx + 4 Γ +z dz. 4 If w st z R θi, θ π, on Γ, thn inquality.9 on th smicircl givs

4 6 SETION 6. +z 4 z 4 R 4. Hnc, by proprty 4., Γ +z 4 dz R 4 (πr). It is clar that th limit of this xprssion is zro as R, and thrfor ( π R ) lim R +x dx + 4 Γ +z dz 4 +x dx. 4 Exampl 6.9 This xampl has illustratd that th contour intgral of /(+z 4 ) around th curv of Figur 6.7 can b usd to valuat th impropr intgral of /( + x 4 ) from ngativ infinity to infinity. Th ral difficulty in such problms is th choic of contour and th choic of intgrand. W now do two mor xampls to illustrat som of ths difficultis. x + dx. Solution W might considr th contour intgral cos z z + dz around th contour in Figur 6.7. rtainly along th straight lin portion of th intgrand rducs to /(x + ), but if w st z x + yi along Γ, thn on th smicircl cos z z + (x+yi)i + (x+yi)i (z +) y+xi + y xi (z +) which bcoms infinit as z. W shall not thrfor b abl to show that th contour intgral along Γ approachs zro as z as in Exampl 6.8. This mans that th choic of cos z/(z + ) was prhaps not a convnint on. onsidr instad z + dz whr is again th contour in Figur 6.7. Sinc th intgrand has simpl pols at z ±i, only z i bing intrior to, ] zi z dz πi Rs + z +,i Thus, π R R xi x + dx + x + dx + i Γ R πi lim z i z + dz sin x x + dx + (z i) zi (z i)(z + i) Γ z + dz ] πi i(i) i π.

5 SETION 6. 6 R x + dx + Γ z + dz (sinc sin x/(x + ) is an odd function). If w st z x + yi R θi on Γ, thn on this smicircl, z + (x+yi)i z (by inquality.9) y R R (sinc y ). Hnc, by inquality 4., Γ z + dz R (πr). Sinc th limit of this xprssion is zro as R, it follows that ( π R ) lim R x + dx + z + dz x dx + Finally, Γ x + dx π. x + dx. Exampl 6. This xampl illustratd that w do not always rplac x s by z s to obtain th appropriat contour intgral. Th following xampl indicats that th choic of contour may not always b obvious. +x dx. Solution Basd on Exampl 6.8 w should prhaps considr +z dz whr is som appropriat contour. larly a part of should b th positiv ral axis and possibly a circular arc of radius R>. But w cannot tak a smicircl as in Exampl 6.8 sinc /( + z ) has a singularity at z p i / G on th ngativ ral axis. What w should lik G f to do thn is choos som othr lin minating from th origin say z r φi, <φ<π, which G R > R z lads to a simpl solution (Figur 6.8). Th Figur 6.8 intgrand has simpl pols at z πi/,, 5πi/. Suppos for th momnt w stipulat that φ b in th intrval π/ <φ<π, but, for th momnt, lav it othrwis arbitrary. auchy s rsidu thorm and L Hôpital s rul giv ] z πi/ dz πi Rs +z +z,πi/ πi lim z πi/ +z

6 6 SETION 6. πi lim z πi/ z Thus, ( i)π Γ +z dz + Γ +z dz + If w st z R θi on Γ, thn on this arc +z z R, and hnc Γ +z dz R (Rφ). Th limit of this xprssion is zro as R. Thus, ( ( i)π lim R Γ +z dz + Γ +z dz + Γ +z dz + +x dx, πi ( i)π. πi/ Γ +z dz. Γ +z dz whr Γ : z r φi, >r (and π/ <φ<π). Our problm thn is to choos φ in ordr that th contour intgral along Γ can b valuatd. Sinc z r φi on Γ, Γ +z dz +r φi φi dr φi If w choos φ π/, thn dz πi/ Γ +z +r dr, and ) +r dr. φi ( i)π πi/ +r dr + ( ) ] i + + +x dx +x dx. Thus, +x dx ( i)π i π. Gnral rsults concrning impropr intgrals of th typs in Exampls ar discussd in Exrciss 7 and.

7 SETION 6. 6 EXERISES 6. In Exrciss 7 us a contour intgral to valuat th dfinit intgral π π π π π π/ π π/ dθ. sin θ dθ sin θ 4 cos dθ 6. θ + sin θ dθ cos θ dθ. +cosθ dθ. sin θ dθ 4. + sin θ dθ 6. 4+cosθ π π π π π/ π/ π/ π + cos θ dθ sin θ + cos θ +4 dθ sin θ + cos θ dθ cos θ + cos θ dθ 6+5sinθ dθ 5 + cos θ dθ 4 + cos θ dθ 5+sinθ dθ cos θ 7. +cosθ dθ In Exrciss 8 us a contour intgral to valuat th impropr intgral dx 9. +x dx. +x6 sin x x dx. +x + π 4. Show that cos n θdθ (n)!π n (n!). 5. (a) Us th substitution u /x to show that +x I +x dx 4 and hnc I 4 (b) Now us contour intgration to calculat I. +x +x 4 dx. x (x + )(x +4) dx x +x dx x (x +9) dx +u +u 4 du, 6. Us contour intgrals to prov th rsult of Exrcis in Sction 4.8.

8 64 SETION Exampl 6.8 and Exrciss 8 ar xampls of impropr intgrals of th form P (x) Q(x) dx whr P (x) and Q(x) ar polynomials (of dgrs m and n), and Q(x) for all ral x. Show that whn n m +, { P (x) sum of th rsidus of P (z)/q(z) at dx πi Q(x) its pols in th half-plan > With this rsult, it is no longr ncssary to introduc th contour of Figur 6.7. Th fact that n m + guarants that th contour intgral along Γ vanishs as R. Us Exrcis 7 to valuat th impropr intgral in Exrciss (x dx 9. +4x +5) }. x + (x + )(x x +) dx. Exampl 6.9 and Exrciss and ar xampls of impropr intgrals of th form P (x) cos ax dx or Q(x) whr P (x) and Q(x) ar polynomials (of dgrs m and n), a> is ral, and Q(x) for any ral x. (a) Us th figur to th right to vrify that sin θ θ π, θ π/, calld Jordan s inquality. Prov that for a>, π ar sin θ dθ π ar ( ar ). (b) Show that whn n m +, and a>, P (x) cos ax dx π Im Q(x) and P (x) sin ax Q(x) y P (x) sin ax Q(x) p/ dx ysin q { sum of th rsidus of P (z) azi /Q(z) at its pols in th half-plan > { sum of th rsidus of P (z) dx πr azi /Q(z) at its pols in th half-plan > Hint: Us th contour in Figur 6.7, and show that R can b chosn sufficintly larg that on Γ, P (z) azi Q(z) ( a m R m ar sin θ + + a ) b n R n, b whr P (z) a m z m + + a and Q(z) b n z n + + b. Now us th rsult in part (a). In Exrciss 5 us th rsult in Exrcis to valuat th impropr intgral. x sin x x. x dx. +5 x 4 +4 dx. sin x x dx 4. + (x + a) + b dx p q } }.

9 π/ SETION ( sin θ) dθ. In Exrciss 6 8 vrify th formula for th givn valus of th paramtrs π π π a + b sin θ dθ a + b cos θ dθ d + a cos θ + b sin θ dθ π, whn < b <a a b π, whn < b <a a b { π sgn d, d > d a b, whr sgn d, d < ar ral with d >a + b., whn a, b, and d

Calculus II (MAC )

Calculus II (MAC ) Calculus II (MAC232-2) Tst 2 (25/6/25) Nam (PRINT): Plas show your work. An answr with no work rcivs no crdit. You may us th back of a pag if you nd mor spac for a problm. You may not us any calculators.